# 应用数学入门 Introduction to Applied Mathematics MAT00003C

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Let $R=K\left[x_{1}, \ldots, x_{n}\right]$ be a polynomial ring over a field $K$ and let $I$ be a monomial ideal of $R$ generated by a finite set of monomials $\left{x^{v_{1}}, \ldots, x^{v_{q}}\right}$. As usual we use $x^{a}$ as an abbreviation for $x_{1}^{a_{1}} \cdots x_{n}^{a_{n}}$, where $a=\left(a_{1}, \ldots, a_{n}\right)$ is in $\mathbb{N}^{n}$. The three central objects of study here are the following blowup algebras: (a) the Rees algebra
$$R[I t]:=R \oplus I t \oplus \cdots \oplus I^{i} t^{i} \oplus \cdots \subset R[t],$$
where $t$ is a new variable, (b) the associated graded ring
$$\operatorname{gr}{I}(R):=R / I \oplus I / I^{2} \oplus \cdots \oplus I^{i} / I^{i+1} \oplus \cdots \simeq R[I t] \otimes{R}(R / I),$$

with multiplication
$$\left(a+I^{i+1}\right)\left(b+I^{j+1}\right)=a b+I^{i+j+1} \quad\left(a \in I^{i}, b \in I^{j}\right),$$
and (c) the symbolic Rees algebra
$$R_{s}(I):=R+I^{(1)} t+I^{(2)} t^{2}+\cdots+I^{(t)} t^{i}+\cdots \subset R[t],$$
where $I^{(i)}$ is the ith symbolic power of $I$.

## MAT00003C COURSE NOTES ：

Let $R=k\left[x_{1}, \ldots, x_{n}\right]$ be a polynomial ring over a field $k$. Suppose $M=x_{1} a_{1} \ldots x_{n} a_{n}$ is a monomial in $R$. Then we define the polarization of $M$ to be the square-free monomial
$$\mathscr{P}(M)=x_{1,1} x_{1,2} \ldots x_{1, a_{1}} x_{2,1} \ldots x_{2, a_{2}} \ldots x_{n, 1} \ldots x_{n, a_{n}}$$
in the polynomial ring $S=k\left[x_{i, j} \mid 1 \leq i \leq n, 1 \leq j \leq a_{i}\right]$.
If $I$ is an ideal of $R$ generated by monomials $M_{1}, \ldots, M_{q}$, then the polarization of $I$ is defined as:
$$P(I)=\left(P\left(M_{1}\right), \ldots, P\left(M_{q}\right)\right)$$

# 实分析 Real Analysis MAT00005C

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{Condition (b) is Riemann’s original formulation of integrability. 1 }
(a) $\Rightarrow$ (b): Assuming $f$ is Riemann-integrable, let
$$\lambda=\int_{a}^{b} f$$

For each $\nu=1, \ldots, n$, choose a sequence $\left(x_{\nu}^{k}\right)$ in $\left[a_{\nu-1}, a_{\nu}\right]$ such that
$$f\left(x_{\nu}^{k}\right) \rightarrow M_{\nu} \text { as } k \rightarrow \infty$$
then
$$\sum_{\nu=1}^{n} f\left(x_{\nu}^{k}\right) e_{\nu} \rightarrow \sum_{\nu=1}^{n} M_{\nu} e_{\nu}=S(\sigma)$$
so by $()$ we have (*)
$$\lambda-\epsilon \leq S(\sigma) \leq \lambda+\epsilon .$$

## MAT00005C COURSE NOTES ：

Similarly,
$$\lambda-\epsilon \leq s(\sigma) \leq \lambda+\epsilon \text {; }$$
thus $S(\sigma)$ and $s(\sigma)$ both belong to the interval $[\lambda-\epsilon, \lambda+\epsilon]$, therefore
$$W_{f}(\sigma)=S(\sigma)-s(\sigma) \leq 2 \epsilon .$$
This proves that $f$ is Riemann-integrable and since
$$S(\sigma) \rightarrow \int_{a}^{b} f \text { as } \mathrm{N}(\sigma) \rightarrow 0,$$
it is clear from $\left({ }^{* *}\right)$ that
$$\lambda=\int_{a}^{b} f . \diamond$$

# 实分析 Real Analysis MATH20101

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conclusion holds for $k=1$. Let us proceed by induction on $k$. If $C$ has no interior, then by Theorem $6.2 .6$ it is included in a hyperplane $H=f^{-1}{c}$ for some non-zero linear form $f$ and $c \in \mathbb{R}$. Then
$$H={x: f(x) \geq c} \cap{x: f(x) \leq c},$$
an intersection of two half-spaces. Also,
$$H=u+V:={u+v: v \in V}$$

Let $V$ be a real vector space and $C$ a convex set in $V$. A real-valued function $f$ on $C$ is called convex iff
$$f(\lambda x+(1-\lambda) y) \leq \lambda f(x)+(1-\lambda) f(y)$$

## MATH20101COURSE NOTES ：

$$t \leq 3 \sum_{i=1}^{q} \lambda\left(J_{i}\right) \leq 6 j \sum_{i=1}^{q} \mu\left(J_{i}\right) \leq 6 j \mu(V) \leq 6 j \varepsilon .$$
Thus $\lambda\left(P_{j}\right) \leq 6 j \varepsilon$. Letting $\varepsilon \downarrow 0$ gives $\lambda\left(P_{j}\right)=0, j=1,2, \ldots$, and letting $j \rightarrow \infty$ proves the lemma.
Now to prove Theorem 7.2.1, for each rational $r$ let
$$g_{r}:=\max (g-r, 0), \quad f_{r}(x):=\int_{a}^{x} g_{r}(t) d t .$$

# 实分析| Real Analysis代写 MT3502

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and $c_{n} \rightarrow a, d_{n} \rightarrow b$; then
$$F\left(d_{n}\right)-F\left(c_{n}\right)=H\left(d_{n}\right)-H\left(c_{n}\right)$$
for all $n$, so in the limit we have
$$F(b)-0=H(b)-0$$

by the continuity of $F$ and In other words,
$$\int_{a}^{b} f=\int_{a}^{b} f \cdot \diamond$$

## MT3502 COURSE NOTES ：

Let $F:[a, b] \rightarrow \mathbb{R}$ and $G:[a, b] \rightarrow \mathbb{R}$ be the indefinite upper integrals of $f$ and $g$, respectively:
$$F(x)=\int_{a}^{-x} f \text { and } G(x)=\int_{a}^{-x} g$$
for $a \leq x \leq b$. If $a<c<d<b$ then, as in the proof of $9.6 .3$,
\begin{aligned} &F(d)=F(c)+\int_{c}^{-d} f, \ &G(d)=G(c)+\int^{-d} g \end{aligned}

# 实分析|Real Analysis  5CCM221A

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Proof. Let $h_{n} \rightarrow 0, h_{n} \neq 0$. Since $E$ is continuous, $E\left(h_{n}\right) \rightarrow E(0)=$ 1. Let $x_{n}=E\left(h_{n}\right)$; then $x_{n}>0, x_{n} \rightarrow 1$ and $x_{n} \neq 1$ (because $\left.h_{n} \neq 0\right)$, so by $9.5 .10$
$$\frac{L\left(x_{n}\right)}{x_{n}-1} \rightarrow 1$$

That is,
$$\frac{h_{n}}{E\left(h_{n}\right)-1} \rightarrow 1$$
whence the lemmn.

## 5CCM221ACOURSE NOTES ：

(i) Show that $A$ is an odd function: $A(-x)=-A(x)$ for all $x \in \mathbb{R}$.
{Hint: $f$ is an even function.}
(ii) $A$ is strictly increasing. {Hint: $\left.A^{\prime}=f .\right}$
(iii) For every positive integer $k$,
$$\frac{1}{1+k^{2}} \leq A(k)-A(k-1) \leq \frac{1}{1+(k-1)^{2}} .$$
{Hint: Integrate $f$ over the interval
(iv) Let
$$s_{n}=\sum_{k=1}^{n} \frac{1}{1+k^{2}} .$$

# 实分析作业代写Real analysis代考

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## 极限 (数学)Limit (mathematics)代写

• 一致收敛Uniform convergence
• 紧空间Compact space
• 连续函数Continuous function
• 一致连续 Uniform continuity

## 实分析的相关

Various ideas from real analysis can be generalized from the real line to broader or more abstract contexts. These generalizations link real analysis to other disciplines and subdisciplines. For instance, generalization of ideas like continuous functions and compactness from real analysis to metric spaces and topological spaces connects real analysis to the field of general topology, while generalization of finite-dimensional Euclidean spaces to infinite-dimensional analogs led to the concepts of Banach spaces and Hilbert spaces and, more generally to functional analysis.

## 实分析课后作业代写

Using the logarithmic function defined in Section $7.1$, the reciprocal of any linear polynomial can be integrated. Indeed, up to a constant multiple, such a function is given by $1 /(x-\alpha)$, where $\alpha \in \mathbb{R}$, and we have
$$\frac{d}{d x}(\ln (x-\alpha))=\frac{1}{x-\alpha} \quad \text { for } x \in \mathbb{R}, x>\alpha$$
The next question that naturally arises is whether we can integrate the reciprocal of a quadratic polynomial, say $x^{2}+a x+b$, where $a, b \in \mathbb{R}$. If this quadratic happens to be the square of a linear polynomial, say $(x-\alpha)^{2}$, then the answer is easy because
$$\frac{d}{d x}\left(\frac{-1}{x-\alpha}\right)=\frac{1}{(x-\alpha)^{2}} \quad \text { for } x \in \mathbb{R}, x \neq \alpha .$$
Further, if the quadratic factors into distinct linear factors, that is, if $x^{2}+a x+b=(x-\alpha)(x-\beta) \quad$ for some $\alpha, \beta \in \mathbb{R}, \alpha>\beta$

# 实分析作业代写Real anlysis代考

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## 代写实分析Real anlysis

• 工业数学Industrial mathematics
• 控制理论Control theory
• 数学方法Mathematical methods
• 几何分析geometric analysis
• 信号处理Signal processing
• 复分析complex analysis
• 傅里叶分析fourier analysis
• 测度论measure theory

## 实分析的历史

1800 年，一段时间以来，因为牛顿卓越的发明 自然哲学的数学原理使数学的光辉照亮了笼罩在假设与猜想的黑暗中的科学。虽然牛顿的思想在当时没有立即被接受，在它出版一个世纪后，“没人可以否认（从《自然哲学的数学原理》中）诞生了一门新的学科，这门学科（至少在特定方面）远远超越了它之前的一切事物，成为科学规范的最佳典范。这就是实分析的雏形微积分。

$$\sum y_{i}\left(x_{i+1}-x_{i}\right),$$

$$y_{j} \leq f(x) \leq y_{j+1}$$

$$\sum m_{j} y_{j},$$
㚳果这些和式当分副加细时䞨向于某个极限, 该极限就是 Lebesgue 意义下的积分. 于是, Lebesgue 称该函数是可和的. 在 1901 年的文章里，Lebesgue 仅棥于讨论 $x$ 和 $y$ 的变化区间有限的情况，稍 后 , 他 去掉了这一约束 .1926 年, 在哥本哈根的一次演汫中, Lebesgue 是这样阐述他的观点的.
“按照 Riemann 的方法, 我们对依自变量 $x$ 的大小肐序所拻供的不可分割的量求和，这 有如没有条理的商人数钱, 碰到硬币数硬币, 碰到纸币数纸币. 而我们的做法像有条理的商人 的做法:

$\cdots \cdots$ 等等. 故, 总共有
$$S=1 \cdot m\left(E_{1}\right)+2 \cdot m\left(E_{2}\right)+5 \cdot m\left(E_{5}\right)+\cdots “$$